AS June 2019 Paper 1 Q2
2 The roots of the equation \(3x^2 - x + 2 = 0\) are \(\alpha\) and \(\beta\).
Find a quadratic equation with integer coefficients whose roots are \(2\alpha - 3\) and \(2\beta - 3\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(y = 2x - 3 \Rightarrow x = \dfrac{y + 3}{2} \Rightarrow 3\left(\dfrac{y + 3}{2}\right)^2 - \dfrac{y + 3}{2} + 2 = 0\) | M1 | 1.1a |
| \(\Rightarrow 3\left(\dfrac{y^2 + 6y + 9}{4}\right) - \dfrac{y + 3}{2} + 2 = 0\) | A1 | 1.1 |
| \(\Rightarrow 3y^2 + 16y + 29 = 0\) | A1 | 1.1 |
| [3] |
Notes
M1: subst \(x = \dfrac{y + 3}{2}\)
or \(x = (1 \pm \sqrt{23}\mathrm{i})/6\)
A1: (1st) expanding bracket correctly
new roots \((-8 \pm \sqrt{23}\mathrm{i})/3\)
A1: (2nd) must have ‘\(=0\)’
B3 if fully correct
Alternative solution
| Scheme | Marks |
|---|---|
| \(\alpha + \beta = 1/3,\ \alpha\beta = 2/3\) \(\Rightarrow 2\alpha - 3 + 2\beta - 3 = 2(\alpha + \beta) - 6 = -16/3\) \((2\alpha - 3)(2\beta - 3) = 4\alpha\beta - 6(\alpha + \beta) + 9 = 29/3\) | M1 |
| New quadratic is \(y^2 + (16/3)y + 29/3 = 0\) | A1 |
| \(\Rightarrow 3y^2 + 16y + 29 = 0\) | A1 |
M1: attempting sum and product of \(2\alpha - 3\) and \(2\beta - 3\)
A1: (1st) sum = −16/3, product = 29/3
A1: (2nd) must have ‘\(= 0\)’